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Word Problems/WORD PROMBLEMS ON SIMULTANEOUS LINEAR EQUATIONS

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Question
HELLO ,
1.WHEN 1 IS ADDED TO THE NUMERATOR OF A FRACTION, THE FRACTION BECOMES 1 BY 2, BUT WHEN 3 IS ADDED TO ITS DENOMINATOR IT BECOMES 1 BY 3 . FIND THE FRACTION .

2.IF 2 IS ADDED TO THE NUMERATOR OF A FRACTION AND 1 IS SUBTRACTED FROM ITS DENOMINATOR , THE FRACTION BECOMES 4 BY 5 .IF 1 IS ADDED TO THE DENOMINATOR OF THE FRACTION , THE FRACTION BECOMES 1 BY 2 . FIND THE FRACTION.

3.THE SUM OF THE DIGITS IN A 2- DIGIT NO. IS 7 .
IF 9 IS SUBTRACTED FROM THE NO. ,ITS DIGITS ARE REVERSED . FIND THE NO.

4.THE DIGIT IN THE TENS PLACE OF A 2- DIGIT NO. IS HALF THE DIGIT IN THE ONES PLACE . IF THE SUM OF THE DIGITS IS 12 , FIND THE NO.

5. A 2-DIGIT NO. OS 5 TIMES THE SUM OF ITS DIGITS. IF 9 IS ADDED TO THE NO. , ITS DIGITS ARE REVERSED . FIND THE NO.

6. A. 2-DIGIT NO. IS 3 TIMES THE THE SUM OF ITS DIGITS. IF 45 IS ADDED TO THE NO. ,ITS DIGITS ARE REVERSED . FIND THE NO.

Answer
1. Solve (a+1)/b = 1/2 and 2a + 2 = b; reverse the 2nd equation into b = 2a + 2,
then put that in the first equation.

2. Solve (a+2)/(b-1) = 4/5 and a/(b+1) = 1/2 by changing the 2nd equation into
a = (b+1)/2, cross multiplying the  1st equation into 5(a+2) = 4(b-1),
and then putting what the 2nd equation was changed to into the 1st equation.

3. Solve (a+2)/(b-1) = 4/5 and a/(b+1) = 1/2 by multiplying the 2nd equation by
(b+1) on both sides and putting this solution for a into the 1st equation.

4. Solve a = b/2 and a+b = 12 by putting the 1st equation solution of a into the 2nd equation.

5. Solve 10a + b = 5(a+b) and 10a + b + 9 = 10b + a by adding -a-b to both sides of the 2nd equation, dividing both sides of this equation by 9 (and seeing what b equals), then putting
this solution for b into the 1st equation and multiplying it out.  Subtract the multiple of 'a' on one side from both sides and the multiples of 'b' on the other side from both sides.
Solve for 'b' and put this solution into the 1st equation.

6. Simplify the two equations 10a + b = 3(a+b) and 10a + b + 45 = 10b + a
for 'a' and 'b on the right side.  From the 1st equation, this should reduce to give
b = 7a/2.  Put this into the 2nd equation and solve for a.

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Scott A Wilson

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I have answered every question that was a story problems that had any relation to math for which an answer existed. I have ever answered some questions which had a vague relation to math, but still related.

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